Multiplying a fraction

  • #1

Homework Statement


I would just like to know if I got this right. ((x2 + 2x + 1)/(18cw3) times ((12c3w)/(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3)(x2 - 18cw3

Homework Equations


((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))

The Attempt at a Solution


(12c3wx2 + 2x12c3w + 12c3w)/((18cw3)(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3x2 - 18cw3
 

Answers and Replies

  • #2
Samy_A
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Removed, as apparently there is an error in the formulation of the question.
 
  • #3
I accidentally used a subscript all the way through instead of a power, argh!!
 
  • #4
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##(12c_3wx_2 + 2x12c_3w + 12c_3w)/(18cw_3x_2 - 18cw_3##
On the right side one of the terms is ##2x12c_3w##. Should that have been ##2x^{12}c^3##? Also, you're missing a right parenthesis.
 
  • #5
haruspex
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((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))
Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.
 
  • #6
On the right side one of the terms is ##2x12c_3w##. Should that have been ##2x^{12}c^3##? Also, you're missing a right parenthesis.
It should have been as haruspex said: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1)
 
  • #7
Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.
Yes, that is what I meant.
 
  • #8
Yes, that is what I meant.
Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.
(x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw3) times (12c3w)/(x-1)
 
  • #9
haruspex
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(x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw3) times (12c3w)/(x-1)
More cancellation to go. Look at the constants, look at the powers of c and w.
 
  • #10
More cancellation to go. Look at the constants, look at the powers of c and w.
(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)
 
  • #13
haruspex
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I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)
You made a mistake with the constants.
 
  • #19
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  • #20
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So the answer is: (x+1)/(6w2) times (4c2)/(x-1)
The constants can be further simplified, and then you should bring it together so you have one fraction.
 
  • #21
The constants can be further simplified, and then you should bring it together so you have one fraction.
Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?
 
  • #22
Samy_A
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Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?
1)
The constants can be further simplified
2) I think you can leave the nominator as ##2c²(x+1)##, and similarly for the denominator. Certainly would give a better readable result.
 
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  • #23
1)

2) I think you can leave the nominator as ##2c²(x+1)##, and similarly for the denominator. Certainly would give a better readable result.
How did you get 2c2(x+1) when it was 4c2/(x+1)?? Isn't the numerator supposed to be 2c2(2x+2)??
And the denominator is supposed to be: 2w2(3x-3)??
 
  • #24
haruspex
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How did you get 2c2(x+1) when it was 4c2/(x-1)?? Isn't the numerator supposed to be 2c2(2x+2)??
And the denominator is supposed to be: 2w2(3x-3)??
You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.
 
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  • #25
You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.
Okay, I think I have it now: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1) = (x+1)2(12c3w)/(18cw2)(x+1)(x-1) Cancel the (x+1) = (12c3w)(x+1)/(18cw2)(x-1) Cancel the c and w = (12c2)(x+1)/(18w)(x-1) divided by 3 (numerator and denominator) = (4c2)(x+1)/(6w)(x-1) divided by 2 (numerator and denominator) = (2c2)(x+1)/(3w)(x-1) Answer
 

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